Evaluate surface integral over surface

[ride4life]

Homework Statement

Evaluate the surface integral of G over the surface S
S is the parabolic cylinder y=2x^2, 0=< x =<5, 0=< z =<5
G(x,y,z)=6x

Answer is one of the following:
1. (15/8)*(401sqrt(401)-1)
2. (5/8)*(401sqrt(401)-1)
3. (15/8)*(401sqrt(401)+1)
4. (5/8)*(401sqrt(401)+1)

Homework Equations

The Attempt at a Solution

Let f(x,y,z)=y-2x^2=0
n=gradf/||gradf||=(-4xi+1j+0k)/sqrt(16x^2+1)
n*=j
G.n=-24x^2/sqrt(16x^2+1)
n.n*=1/sqrt(16x^2+1)
therefore the double integral over S = SS (G.n/n.n*) dzdx
solving the double integral gets -5000

 

[algebrat]

Let me try ot reason through what you have. If at the end you were integrating dzdx, I'll assume I should try to paramet(e)rize the surface in x and z.

Then r(x,z)=(x,2x^2,z). Then r_x=(1,4x,0), while r_z=(0,0,1).

Then cross product is (r_x)x(r_z)=(4x,-1,0).

Then the infinitesimal area on the surface is given by

dS=sqrt(16x^2+1)dxdz.

Now we want to integrate the SCALAR G(x,y,z)=4x against the area.

That is, integrate G dS.

So $\int_{x=0}^5\int_{z=0}^54x\sqrt{16x^2+1}\ dzdx.$

So my first guess is, you are mixing up VECTOR integrals with SCALAR integrals. You might compare them and their derivations.